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Wewaii [24]
3 years ago
14

What is the molarity of a 250.0 milliliter aqueous solution of sodium hydroxide that contains 15.5 grams of solute? 6.2 M

Chemistry
2 answers:
Alchen [17]3 years ago
8 0
Moles = 15.5 g / 40 g/mol = 0.3875 mol

M = 0.3875 mol / 0.250 L = 1.55M
Digiron [165]3 years ago
6 0

Answer : The molarity is 1.55 M

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

or,

\text{Molarity}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

In this question the solute is, sodium hydroxide.

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{15.5g\times 1000}{40g/mole\times 250.0L}

\text{Molarity}=1.55mole/L=1.55M

Therefore, the molarity is 1.55 M

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Answer:

a) ΔHvap=35.3395 kJ/mol

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Explanation:

Given the reaction:

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Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

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The Clasius-Clapeyron equation is:

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ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

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T₁=323.15K

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ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

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